Two dice are thrown simultaneously. If \( X \) denotes the number of sixes appearing on two dice, then which of the following are correct?
(A) \( P(X=0)=\displaystyle\frac{1}{36} \)
(B) \( P(X=1)=\displaystyle\frac{10}{36} \)
(C) \( P(X=1)+P(X=0)=\displaystyle\frac{35}{36} \)
(D) \( P(X=1)+P(X=2)=\displaystyle\frac{11}{36} \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (B), (C) and (D) only
On each die, P(six) = \( \frac{1}{6} \) and P(not six) = \( \frac{5}{6} \). The six in X = 1 can be on either die, hence the factor 2.
\( \displaystyle P(X=0)=\frac{5}{6}\times\frac{5}{6}=\frac{25}{36} \)
\( \displaystyle P(X=1)=2\times\frac{1}{6}\times\frac{5}{6}=\frac{10}{36} \)
\( \displaystyle P(X=2)=\frac{1}{6}\times\frac{1}{6}=\frac{1}{36} \)
So \( P(X=0)=\frac{25}{36} \), \( P(X=1)=\frac{10}{36} \) and \( P(X=2)=\frac{1}{36} \).
Checking the statements
(A) \( P(X=0)=\frac{1}{36} \) — Incorrect. \( P(X=0)=\frac{25}{36} \); \( \frac{1}{36} \) is \( P(X=2) \).
(B) \( P(X=1)=\frac{10}{36} \) — Correct. \( 2\times\frac{1}{6}\times\frac{5}{6}=\frac{10}{36} \).
(C) \( P(X=1)+P(X=0)=\frac{35}{36} \) — Correct. \( \frac{10}{36}+\frac{25}{36}=\frac{35}{36} \).
(D) \( P(X=1)+P(X=2)=\frac{11}{36} \) — Correct. \( \frac{10}{36}+\frac{1}{36}=\frac{11}{36} \).